Today’s reagent is one that most students have experience in making at some point or another. Grignard reagents are formed by the reaction of magnesium metal with alkyl or alkenyl halides. They’re extremely good nucleophiles, reacting with electrophiles such as carbonyl compounds (aldehydes, ketones, esters, carbon dioxide, etc) and epoxides. They’re also very strong bases and will react with acidic hydrogens (such as alcohols, water, and carboxylic acids).

Similar to or the same as: very similar to organolithium reagents.

Examples:

Grignard reagents are made through the addition of magnesium metal to alkyl or alkenyl halides. The halide can be Cl, Br, or I (not F). It’s slightly easier to make Grignards from the iodides and bromides, however. Note what’s happening here – the magnesium is “inserting” itself between the carbon and the halide. This halide the “X” referred to when we refer to Grignard reagents as “RMgX”.

One of the most common uses of Grignard reagents is in their reaction with aldehydes and ketones to form alcohols. In the first step, the Grignard forms the carbon-carbon bond. This results in an alkoxide (the conjugate base of an alcohol). To form the alcohol, it’s necessary to add acid at the end of the reaction (in what’s called the “workup” step). This is shown here as “H3O+” (the “X” is just the counter-ion, a spectator here)

The reaction behaves similarly with ketones. Again, there’s nothing special about the Cl here – it all depends on how you made the Grignard in the first place.

Grignard reagents will also add to esters. What makes these reactions a little more complicated is that they add twice. The net result (after addition of acid) is a tertiary alcohol. This is also the case for acid halides (acyl halides) and anhydrides. One notable exception is carboxylic acids (more on that below).

Another important reaction of Grignard reagents is that they will add to epoxides to form carbon-carbon bonds. One thing to keep in mind here is that the tendency is for them to add to the less substituted end of the epoxide – that is, the less sterically hindered end. You can think of this reaction as being essentially similar to an SN2 reaction. After addition of acid, an alcohol is obtained.

Grignard reagents also add to carbon dioxide (CO2) to form carboxylates, in a reaction similar to their reactions with ketones and aldehydes. The carboxylates are converted to carboxylic acids after addition of acid (such as our trusty H3O(+) ).

Finally, since Grignard reagents are essentially the conjugate bases of alkanes, they’re also extremely strong bases. This means that sometimes acid-base reactions can compete with their nucleophilic addition reactions. One common situation where this crops up is when Grignard reagents are added to carboxylic acids. It’s easy to forget that carboxylic acids… are acids. This means that instead of adding to the carbonyl, they react with the proton instead and form the carboxylate salt.

This can also be used to convert alkyl halides to alkanes. First you treat it with magnesium, and then you treat the Grignard with a strong acid. This gives you the alkane. You can also use this to introduce deuterium (D) into molecules! The first step is to make the Grignard reagent. The second is to treat that Grignard with a deuterated acid such as D2O. This gives you the deuterated alkane!

So how does it work? The key to the Grignard reagent is actually very simple. When you think about the relative electronegativities of carbon (2.5) and magnesium (1.1), the bond between carbon and magnesium is polarized toward carbon. That means that carbon is more electron rich than magnesium and is actually nucleophilic! Here’s a closer look.

In the reaction of Grignards with aldehydes, the carbon attacks the carbonyl carbon and performs a 1,2-addition to give an alkoxide. In the second step, acid is added to give you the alcohol.

There are so many other elements to the Grignard but a limited amount of space. So I’ll leave it there. If you want more details you’ll have to check out the Reagent Guide!

P.S. You can read about the chemistry of Grignard reagents and more than 80 other reagents in undergraduate organic chemistry in the “Organic Chemistry Reagent Guide”, available here as a downloadable PDF. The Reagents App is also available for iPhone, click on the icon below!

Comments

I recently came across the interesting set of reactions between Grignard reagents and (terminal) propargylic chlorides. SN2′ substitution gives allenes with a new C-C single bond, along with a mess of other stuff resulting from deprotonation of the terminal alkyne…including the venerable alkylidene carbene, perhaps my favorite obscure intermediate in all of organic chemistry!

“Greg Nard added an equivalent of 3,4-epoxy-4-methylcyclohexanol (Figure 1) to an ether solution of methyl magnesium bromide and then added dilute hydrochloric acid. He expected that the product would be a diol (Figure 2). He did not get any of the expected product. What product did he get?” i have tried everything for the answer i cannot figure it out

Sounds like it just deprotonated the hydroxyl group at the 1 position in the ring, assuming your substrate is 3,4-epoxy-4-methylcyclohexan-1-ol. That alkoxide could probably then go open that epoxide to make a product I can’t name, but should look like a dicyclohexylether with the methyl and erstwhile epoxide OH on one of the rings alpha to the ether linkage. So no diol from having the Grignard open the epoxide because the -OH is too acidic a proton to have in the presence of the CH3MgBr.

Ooooh, or you might get a cool intramolecular attack if you throw your substrate in a chair conformation an let the O- attack the epoxide that way. I don’t know. It’s worth building a model or sketching it out. If you were running a very dilute solution of substrate such that two of those molecules finding each other were a very rare prospect, the intramolecular possibility would win.

In the intramolecular case, though, you’d end up opening the epoxide at the more substituted side to get two 5-membered rings sharing the C-O-C bridge (gotta get the model out for this one!) instead of 4+6…I think…

Great question. It tends to initiate electron-transfer type reactions that end up leading to cleavage of the R-X (X being halogen) bond. Grignards tend to be clusters in solution and so they are more sterically hindered than they might appear. The solution is to make them into organocuprates (Gilman reagents) by using CuBr or the like. Then SN2 reactions work well (particularly on primary substrates). James

We recently did a grignard synthesis in lab using diethyl ether as the solvent and it said that the signs of reaction were cloudiness of the precipitate and bubbles showing up on the magnesium but I can’t figure out what kind of gas these bubbles would be…Would they be the diethyl ether?

Same thing! Water is a strong enough acid to protonate the alkoxide. [Extra detail, feel free to skip: this will form HO- and ROH . Why doesn’t the reaction just go in reverse, since ROH is of similar acid strength to water? The key is to add a large excess of water. Since there will be a much larger concentration of water, equilibrium drives it forward to the alcohol.]

The short answer is no. The electons in a C-C pi bond in an alkene are shared relatively equally between the carbons, with the result that neither carbon has any significant partial positive charge. Alkenes are unreactive. Contrast that to a C=O bond, where carbon is partially positive and oxygen is partially negative; Grignard reagents are much more reactive in that situation.